Proceed by <mathematical induction> on $n$. The cases $n=0,1$ are immediate. Given a <symmetric chain decomposition of a Boolean lattice> $\mathcal P([n-1])$, consider one of its chains
$$
A_r\subset A_{r+1}\subset\cdots\subset A_{n-1-r},
\qquad |A_j|=j.
$$
It produces the two chains
$$
A_r\subset\cdots\subset A_{n-1-r}
\subset A_{n-1-r}\cup\{n\}
$$
and, when nonempty,
$$
A_r\cup\{n\}\subset A_{r+1}\cup\{n\}
\subset\cdots\subset A_{n-2-r}\cup\{n\}.
$$
The first runs from rank $r$ to rank $n-r$, and the second from rank $r+1$ to rank $n-1-r$; both endpoint ranks sum to $n$. They are disjoint and together contain the old chain both without and with $n$. Doing this for every old chain partitions $\mathcal P([n])$ into symmetric chains.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 109 / 1 / a / Solution

A symmetric chain in the Boolean lattice $\mathcal P([n])$ contains one set of each size from $r$ through $n-r$. A symmetric chain decomposition partitions the Boolean lattice into such chains.


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